Video.

They describe subgroups $H \subset G$, $G$ finite group. where

$|H| = p^m$, $p$ a prime.

First Sylow theorem

Gives partial converse to Lagrange's theorem

For $p$ a Prime number

$|G| = p^m r$, where $p$ does not divide $r$.

Then there is always a Subgroup with $p^m$ elements. Such subgroup is called Sylow p-group.

Precise statements.

A corollary is Cauchy's theorem

Corollary. Let p be a prime, and G a group of order $p^2$. Then either $G \cong Z/p^2 Z$ or $\cong Z/pZ \times Z/pZ$ ($Z$ Integers). See Direct product of groups.

Remember already know that if the order of a group is prime^2, then the group is commutative (shown using Class equation). A particular example of a p-group.

Theorem. Let $p$ be a prime, let $G$ be a group of order $2p$. Then either $G \cong Z/2pZ$ (cyclic) or $G \cong D_p$ (Dihedral group). Proof

Proof of first Sylow theorem. Uses some combinatorics, and the left multiplication action acting on subgroups of size $p^m$.

Second Sylow theorem

In particular it says, all Sylow subgroups are Conjugated to each other.

Let subgroup $K \subset G$, whose order is divisible by $p$. Let $H$ be a Sylow p-subgroup of $G$. Then there is a conjugate subgroup $H' = gHg^{-1}$ of $G$, such that $K \cap H'$ is a p-Sylow subgroup of $K$.

Corollaries.

1. Let $K \subset G$ be a subgroup which is a p-group. Then $K$ is contained in a p-Sylow subgroup of $G$.
2. The Sylow p-subgroups of $G$ are all conjugate

As conjugation is bijection, the conjugate of a Sylow p-subgroup is a Sylow p-subgroup.

Proof of second Sylow theorem

Third Sylow theorem

Tells you something about how many Sylow p-subgroups. $|G| = p^m r$, $p \nmid r$ (does not Divide). Let $s$ be the number of Sylow p-subgroups. Then $s$ divides $r$, and $s \cong 1 \text{ mod } p$.

Theorem. Let $G$ be a group of order $pq$, $p >q$ both Prime numbers. If $q$ does not divide $p-1$, then $G$ is cyclic.

Proof of third Sylow theorem

Partial classification of finite groups of small order. Classified all groups up to order 10 (except 8)